Power Tool Battery Charger Question

[fishman94]

My battery charger recently quit working for one of my drills, so I decided to take it apart to see why and found the 120->~11VAC transformer wasn't outputing any voltage. I assumed the transformer got too hot one time and burnt out the temperature fuse sealed into the windings.

Anyway, the circuit used looked pretty simple, but I'm trying to figure out how exactly the paralleled resistor and resistor-LED work in the circuit (attached).

Its a simple bridge rectifier before that and the charger is rated for 10.15 V at 215mA and 11.3 V at 120mA, and my calculations seems to show a larger resistor is needed to limit the current to those values (but thats excluding the LED part that is confusing me - and the LED only comes on when a battery is instlled into the charger.
Does the battery charge somehow affect the LED?

Thanks,
Arthur

 

[kchriste]

What is the nominal battery voltage?

 

[Hero999]

The transformer is probably 9V, it's just slightly higher when no load is connected.

The LED is just there to show it's charging.

When no battery is connected the voltage across the 5R resistor is zero so the LED is off.

When a battery is connected some current flows through the 5R resistor so a voltage appears across it so the LED turns on.

The 100R resistor limits the current through the LED to a safe level.

I suggest you measure the resistor again, it's probably higher than 5R, with 215mA flowing the voltage will only be 1.075V which isn't enough to turn on an LED.

 

[3lectrokid]

you could have a look at this battery charger if you want to build your own: NiCd Battery Charger

 

[fishman94]

The battery is a 9.6 V NiCd. If I measure the voltage on the output where the battery connects at, it shows about 10 V without the battery installed (and measuring it doesn't turn the led on). I have checked the 5R resister multiple times because I thought it should be higher - and also taken apart a working charger (a 14.4V one though) and the values are the same for that charger.
From what I'm taking of what you said, the 5R causes a small voltage drop to create a voltage difference across the 100R and led, correct? Also then, where does the 215mA get calculated from - since that would require ~50R at around the given voltage, and computing the two resistors as if they were in parallel is less than 5R. How does adding the LED in affect the calculation?

The transformer is probably 9V, it's just slightly higher when no load is connected.

The LED is just there to show it's charging.

When no battery is connected the voltage across the 5R resistor is zero so the LED is off.

When a battery is connected some current flows through the 5R resistor so a voltage appears across it so the LED turns on.

The 100R resistor limits the current through the LED to a safe level.

I suggest you measure the resistor again, it's probably higher than 5R, with 215mA flowing the voltage will only be 1.075V which isn't enough to turn on an LED.

 

[Hero999]

The current depends on the battery and transformer voltage and can be calculated using Ohm's law.

What does it say on the resistor? Perhaps it's damage and the resistance has changed?

 

[fishman94]

The resistor was darkened so I couldn't make out the colors too well - and the charger that still works was worse for apparent damage (same exact circuit) and still read just 5Ωs.
That basically means the resistors in the circuit have nothing to do with the current that charges the battery, but rather simply to light the led then, right?

The current depends on the battery and transformer voltage and can be calculated using Ohm's law.

What does it say on the resistor? Perhaps it's damage and the resistance has changed?

 

[Hero999]

The resistor limits the current to the battery, without it, the battery will overheat and could explode.

From the sounds of it, the resistor is underated and should be replaced with a higher wattage rating even if it hasn't yet failed.

Have you measured the resistance of the resistor in the working charger?

The resistor failing could be the reason for the transformer burning out: failing to limit the current probably could cause the transformer to overheat and luckily it failed before the batteries.

 

[fishman94]

Yes, I have measured the resistance in the working charger - it was also only 5Ohms. How would Kirkoff's rules apply with an led inline with one of the paralleled resistors - I can figure it out without theled, but with it do you just take a voltage drop with it?

 

[Diver300]

The current through the LED is so small that it does not significantly affect the charging current. Also you can work out the voltage drop on the 5 ohm resistor and then see how much current that voltage would give in the LED.

With no smoothing capacitor the current will have a large AC component. Although a voltage drop of 1 V won't light the LED, the peak voltage drop will be much larger, so the LED will light. Certainly, the 12V Lead-Acid charger that I have has an LED that flickers at 100 Hz

 

[Hero999]

Then the current must be higher than 215mA to light it.

Assuming a forward voltage of 1.8V, the current should be at least 360mA to light the LED, because it's a rectifier, the current will be pulsed so the LED will light with RMS current of 255mA, 360mA peak.

 

[Diver300]

Hero, I think you have a slightly too simple view of the current waveform.

If you have a purely resistive load on a full-wave rectified sine wave, the peak current will be sqrt(2) times the RMS current, which I think is where you got the 360 / 255 ratio.

However the RMS current isn't important for charging a battery, it's the average current. Also the peak current will be much larger on a full-wave rectified transformer charging a battery. The transformer voltage will be less than the battery voltage for about 50% of the time, so all the current will be in the other half of the time.

I think that the peak current could well be 3 times the average current, so the LED will probably light at about 120 mA average.

 

[Hero999]

You're right, I forgot the current will be complex waveform rather than a simple sinewave.

If the battery voltage is 9.6V and the transformer peak voltage is 15V, the peak current will be 1.08A and the peak voltage across the LED and resistor will be 5.4V so the peak current will be 36mA.